Question

A short bar magnet produces a magnetic field of \(6.4 \times 10^{-5}\) T at a distance of 20 cm from the center of the magnet on the normal bisector of the magnet. The magnetic field produced by this magnet at a distance of 40 cm on the axis is

• A. \(4.8 \times 10^{-5}\) T
• B. \(3.2 \times 10^{-5}\) T
• C. \(1.6 \times 10^{-5}\) T
• D. \(6.4 \times 10^{-5}\) T

Hint:

Magnetic field due to bar magnet decreases as \(1/r^3\).
Axial field is double the equatorial at same distance.

Answer 1

The Correct Option is C

Field on equatorial line: \[ B_{\text{eq}} = \frac{\mu_0}{4\pi} \cdot \frac{M}{r^3} \] Field on axial line: \[ B_{\text{axial}} = \frac{\mu_0}{4\pi} \cdot \frac{2M}{r^3} \] Given \( B_{\text{eq}} = 6.4 \times 10^{-5} \) T at \( r = 0.2 \) m. The ratio of axial to equatorial field is: \[ \frac{B_{\text{axial}}}{B_{\text{eq}}} = 2 \left( \frac{r_{\text{eq}}}{r_{\text{axial}}} \right)^3 = 2 \left( \frac{20}{40} \right)^3 = 2 \cdot \frac{1}{8} = \frac{1}{4} \] \[ B_{\text{axial}} = \frac{1}{4} \cdot 6.4 \times 10^{-5} = 1.6 \times 10^{-5} \text{ T} \]